| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve |linear| > |linear| |
| Difficulty | Moderate -0.3 Part (i) is straightforward identification of transformations (horizontal translation and horizontal stretch). Part (ii) requires squaring both sides to eliminate moduli, then solving a quadratic inequality—a standard technique for C3 level. The working is methodical rather than requiring insight, making this slightly easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Refer to translation and stretch | M1 | In either order; ignore details here; allow any equiv wording (such as move or shift for translation) to describe geometrical transformation but not statements such as add 3 to \(x\) |
| Either: State translation in negative \(x\)-direction by 3 | A1 | Or state translation by \(\begin{pmatrix}-3\\0\end{pmatrix}\); accept horizontal to indicate direction; term 'translate' or 'translation' needed for award of A1 |
| State stretch by factor 2 in \(y\)-direction | A1 | Or parallel to \(y\)-axis or vertically; term 'stretch' needed for award of A1; these two transformations can be given in either order. SC: if M0 but details of one transformation correct, award B1 for 1/3 (in Either, Or 1, Or 2 cases) |
| Or 1: State stretch by factor \(\frac{1}{2}\) in \(x\)-direction | A1 | Or parallel to \(x\)-axis; term 'stretch' needed for award of A1 |
| State translation in negative \(x\)-direction by 3 | A1 | Or state translation by \(\begin{pmatrix}-3\\0\end{pmatrix}\); term 'translate' or 'translation' needed; these two transformations must be in this order – if details correct for M1A1A1 but order wrong, award M1A1A0 |
| Or 2: State translation in negative \(x\)-direction by 6 | A1 | Or state translation by \(\begin{pmatrix}-6\\0\end{pmatrix}\); term 'translate' or 'translation' needed |
| State stretch by factor \(\frac{1}{2}\) in \(x\)-direction | A1 | Or parallel to \(x\)-axis; term 'stretch' needed for award of A1; these two transformations must be in this order – if details correct for M1A1A1 but order wrong, award M1A1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Either: Solve linear eqn/ineq to obtain critical value \(-6\) | B1 | |
| Attempt solution of linear eqn/ineq where signs of \(x\) and \(2x\) are different | M1 | |
| Obtain critical value \(-2\) | A1 | |
| Attempt solution of inequality | M1 | Using table, sketch, …; implied by correct answer or answer of form \(a |
| Obtain \(-6 < x < -2\) | A1 | As final answer; must be \(<\) not \(\leq\); allow "\(x>-6\) and \(x<-2\)" |
| Or: Square both sides to obtain \(x^2 > 4(x^2+6x+9)\) | B1 | Or equiv |
| Attempt solution of 3-term quadratic eqn/ineq | M1 | With same guidelines as in Q2(ii) for factorising and formula |
| Obtain critical values \(-6\) and \(-2\) | A1 | |
| Attempt solution of inequality | M1 | Using table, sketch, …; implied by correct answer or answer of form \(a |
| Obtain \(-6 < x < -2\) | A1 | As final answer; must be \(<\) not \(\leq\); allow "\(x>-6\) and \(x<-2\)" |
## Question 5:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Refer to translation and stretch | M1 | In either order; ignore details here; allow any equiv wording (such as move or shift for translation) to describe geometrical transformation but not statements such as add 3 to $x$ |
| **Either:** State translation in negative $x$-direction by 3 | A1 | Or state translation by $\begin{pmatrix}-3\\0\end{pmatrix}$; accept horizontal to indicate direction; term 'translate' or 'translation' needed for award of A1 |
| State stretch by factor 2 in $y$-direction | A1 | Or parallel to $y$-axis or vertically; term 'stretch' needed for award of A1; these two transformations can be given in either order. SC: if M0 but details of one transformation correct, award B1 for 1/3 (in Either, Or 1, Or 2 cases) |
| **Or 1:** State stretch by factor $\frac{1}{2}$ in $x$-direction | A1 | Or parallel to $x$-axis; term 'stretch' needed for award of A1 |
| State translation in negative $x$-direction by 3 | A1 | Or state translation by $\begin{pmatrix}-3\\0\end{pmatrix}$; term 'translate' or 'translation' needed; these two transformations must be in this order – if details correct for M1A1A1 but order wrong, award M1A1A0 |
| **Or 2:** State translation in negative $x$-direction by 6 | A1 | Or state translation by $\begin{pmatrix}-6\\0\end{pmatrix}$; term 'translate' or 'translation' needed |
| State stretch by factor $\frac{1}{2}$ in $x$-direction | A1 | Or parallel to $x$-axis; term 'stretch' needed for award of A1; these two transformations must be in this order – if details correct for M1A1A1 but order wrong, award M1A1A0 |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| **Either:** Solve linear eqn/ineq to obtain critical value $-6$ | B1 | |
| Attempt solution of linear eqn/ineq where signs of $x$ and $2x$ are different | M1 | |
| Obtain critical value $-2$ | A1 | |
| Attempt solution of inequality | M1 | Using table, sketch, …; implied by correct answer or answer of form $a<x<b$ or of form $x<a, x>b$ (where $a<b$); allow $\leq$ here |
| Obtain $-6 < x < -2$ | A1 | As final answer; must be $<$ not $\leq$; allow "$x>-6$ and $x<-2$" |
| **Or:** Square both sides to obtain $x^2 > 4(x^2+6x+9)$ | B1 | Or equiv |
| Attempt solution of 3-term quadratic eqn/ineq | M1 | With same guidelines as in Q2(ii) for factorising and formula |
| Obtain critical values $-6$ and $-2$ | A1 | |
| Attempt solution of inequality | M1 | Using table, sketch, …; implied by correct answer or answer of form $a<x<b$ or of form $x<a, x>b$ (where $a<b$); allow $\leq$ here |
| Obtain $-6 < x < -2$ | A1 | As final answer; must be $<$ not $\leq$; allow "$x>-6$ and $x<-2$" |
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5 (i) Give full details of a sequence of two transformations needed to transform the graph of $y = | x |$ to the graph of $y = | 2 ( x + 3 ) |$.\\
(ii) Solve the inequality $| x | > | 2 ( x + 3 ) |$, showing all your working.
\hfill \mbox{\textit{OCR C3 2013 Q5 [8]}}